# Number Theory Proof

If p is a prime number and k≥1, then there are exactly (p^k)/p numbers between 1 and (p^k) that are divisible by p. Could you please prove this or provide some links for solution.

This should be easy to prove with basic maths.

Let S be the set of numbers divisible by p in range [1, p^{k}] where k >=1 are :

Mathematically,
S = \{ x : p | x , and\space x\ \in\ [1, p^k] \}

\therefore S = \{ p, 2p,3p,4p,...p^k\} or
\therefore S = \{ p*i : i \in\ [1, p^{k-1}]\}

Thus, total terms in set S are simply the number of multiples of p in [1, p^k] which is p^{k-1}.

Alternatively,
Set S forms an Arithmetic Progression with,
a = p,
d = p,
t_n = p^k

Using n^{th} term formula,
t_n = a + (n - 1)*d
Substituting, above values, we get,
p^k= p + (n - 1)*p
\therefore n = p^{k-1}

As pointed out by @everule1, p can be any natural number.

6 Likes

Shouldn’t this be true for all numbers?

2 Likes

Yes, you are right. But since the OP mentioned prime, I considered primes at first thought. But p can be really any natural number.
I have corrected the claim.